Entire solutions for a class of variational problems involving the biharmonic operator and Rellich potentials

Introduction

In this paper we study weak solutions to problem

Our main assumptions are the following: $N\geq 5$ and

It is well known that $\gamma_{N}^{2}$ is the best constant in the Rellich inequality

Some existence and multiplicity results for (1.1) in case $\lambda=0$ can be found in . If $q=2^{*\!*}$, where

is the critical Sobolev exponent, then (1.1) becomes

The second-order version of (1.5), namely,

was studied by Terracini , under the assumption $\lambda<(N-2)^{2}/4$, the Hardy constant. We quote also , and references there-in for a larger class of problems related to the Caffarelli-Kohn-Nirenberg inequalities .

In the first part of the paper we deal with radially symmetric solutions to (1.1). By using the Rellich inequality (1.4) and the results in one can check that for any $q,\lambda$ and $\beta$ satisfying (1.2), the infimum

is positive. The following existence result holds.

Let $N\geq 5$, $q>2$, $\lambda<\gamma_{N}^{2}$, and put $\beta=N-q(N-4)/2$. Then problem (1.1) has at least one radially symmetric solution $u$ that achieves $S_{q}^{\rm rad}(\lambda)$.

Notice that problem (1.1) is invariant with respect to the weighted dilation

From now on we will identify solutions that coincide up to a weighted dilation and change of sign. This agreement is used in the next uniqueness and positivity result.

Let $N\geq 5$, $q>2$, and put $\beta=N-q(N-4)/2$. If

Theorems 1.1 and 1.2 will be proved in Section 3. Some preliminary results of independent interest will be proved in Section 2.

Notice that the positivity of $u$ does not follow from its characterization as an extremal for $S_{q}^{\rm rad}(\lambda)$. Indeed, in problems involving the biharmonic operator a breaking positivity phenomenon might appear and ground state solutions might be forced to change sign, see for instance . Actually one could wonder whether extremals for $S_{q}^{\rm rad}(\lambda)$ change sign if $\lambda<<0$. A similar phenomenon would be completely new with respect to (1.6) and to similar second-order problems.

In case $q\leq 2^{*\!*}$ one can use again variational methods to find solutions to (1.1) that are not necessarily radially symmetric. We recall that

where $S^{*\!*}$ is the Sobolev constant. If $q$ and $\beta$ are as in (1.2), then by interpolating (1.8) and (1.4) via Hölder inequality, one plainly gets that there exists a constant $C=C(N,q)>0$ such that

Notice that (1.9) is the second-order version of the celebrated Caffarelli-Kohn-Nirenberg inequalities ; we quote also for a large class of dilation-invariant inequalities on cones. It is clear that the infimum

is positive, provided that $\lambda<\gamma_{N}^{2}$. In addition, extremals for $S_{q}(\lambda)$ give rise to solutions to (1.1). In Section 4 we prove the next existence result.

Let $N\geq 5$, $q\in(2,2^{*\!*}]$, $\lambda<\gamma_{N}^{2}$, and put $\beta=N-q(N-4)/2$.

The infimum $S_{q}(\lambda)$ is achieved for any $q\in(2,2^{*\!*})$.

The infimum $S_{2^{*\!*}}(\lambda)$ is achieved if and only if $\lambda\geq 0$.

Clearly enough, it results that $S_{q}(\lambda)\leq S_{q}^{\rm rad}(\lambda)$. In Section 5 we wonder whether the solutions in Theorem 1.3 are radially symmetric or breaking symmetry occurs. First, by using rearrangement techniques we prove that $S_{q}(\lambda)=S_{q}^{\rm rad}(\lambda)$ provided that $\lambda\geq 0$. In contrast, we show that if $\lambda<<0$ then $S_{q}(\lambda)<S_{q}^{\rm rad}(\lambda)$. Therefore, if in addition $q\in(2,2^{*\!*})$ then problem (1.1) has at least two distinct solutions.

Homoclinic solutions to a fourth order ODE

To prove Theorems 1.1 and 1.2 we need some preliminary results that hold for ordinary differential equations involving differential operators of the form

We start with a lemma about a linear equation.

Proof. If $\eta^{\prime\prime\prime}(0)=0$ then we are done, since we immediately infer $\eta\equiv 0$. We argue by contradiction. Replacing $\eta$ with $-\eta$ if needed, we can assume that $\eta^{\prime\prime\prime}(0)>0$. Since $A\geq B$, then the equation $c^{2}-2A~{}\!c+B^{2}=0$ has two real nonnegative roots $c_{+},c_{-}$. Next we put

Since $6\eta(s)=\eta^{\prime\prime\prime}(0)s^{3}+o(s^{3})$ and $\psi(s)=-\eta^{\prime\prime\prime}(0)s+o(s)$ as $s\to 0^{+}$, and since $\eta^{\prime\prime\prime}(0)>0$, then $\eta>0$ and $\psi<0$ in a right neighborhood of . In addition, $\eta$ is smooth and vanishes at infinity. Thus there exists $s_{\!\infty}\in(0,\infty]$ such that $\eta>0$ in $(0,s_{\!\infty})$ and $\eta\in H^{1}_{0}(0,s_{\!\infty})$. We claim that $\psi<0$ on $(0,s_{\!\infty})$. If not, there exists $s_{1}\in(0,s_{\!\infty})$ such that $\psi<0$ on $(0,s_{1})$ and $\psi\in H^{1}_{0}(0,s_{1})$. But then from (2.3) we readily get

which is impossible. Finally, using $\eta\in H^{1}_{0}(0,s_{\!\infty})$ as test function in (2.2) and recalling that $\eta>0$ and $\psi<0$ on $(0,s_{\!\infty})$, we get

a contradiction. The Lemma is completely proved. $\square$

Next we state the main result of this section.

Let $q>2$ and let $A,B$ be two given constants, such that $A,B>0$.

Next, we assume that $A\geq B$. We first notice that $w$ has at least one critical point, since it is smooth and vanishes at infinity. The next claim is the main step in our argument:

To prove the claim, we first notice that we can assume that $s_{0}=0$. We put

and we put $a(s)\equiv 0$ on $\{\eta^{-1}(0)\}$. Notice that

then the unique solution to (2.4) is given by

On the other hand, it has been proved in that a plethora (countably infinite set) of sign-changing homoclinic solutions exists for $A/B<1$ close enough to $1$.

The differential equation in (2.4) is conservative, with Hamiltonian

Therefore, any homoclinic solution $w$ to (2.4) satisfies

Using (2.8) one infers the following a-priori bound on homoclinic solutions to (2.4):

Radially symmetric solutions

In this section we prove Theorems 1.1 and 1.2. The main tool is the Emden-Fowler transform, which has already been used in .

Given a radially symmetric function $u$ we define the Emden Fowler transform of $u$ by $w=Tu$, where

Proof of Theorem 1.1. From (3.1) and (3.2) it immediately follows that $u$ achieves $S^{\rm rad}_{q}(\lambda)$ if and only if $w=Tu$ attains the infimum $I(\gamma_{N}+2,\gamma_{N}^{2}-\lambda)$, see (2.5). The conclusion follows from $(i)$ of Theorem 2.2. $\square$

The conclusion readily follows from the second part of Theorem 2.2. $\square$

Using Remark 2.3, one can explicitly compute the radially symmetric solution to (1.1) when

Notice that $\lambda(q)>-(N-2)^{2}$ for any $q\in(2,2^{*\!*}]$ and that $\lambda(2^{*\!*})=0$. If $\lambda=\lambda(q)$ then the unique radially symmetric solution to (1.1) is given by

for a computable constant $\widetilde{C}$ depending on $N,q$, where

Let $N\geq 2$ be any integer, and let $\alpha$ be a given exponent, such that

By the results in , , it turns out that for any $q\in[2,\infty)$ the infimum

More precisely, for $q=2$ the radial Rellich constant is given by

We point out the following positivity and uniqueness result.

Proof. Existence follows from Theorem 2.8 in . Following , we introduce the value

Here we have denoted by $w=T_{\alpha}u$ the weighted Emden-Fowler transform of $u$:

Since $\overline{\gamma}_{N\!,\alpha}\geq\gamma_{N\!,\alpha}$, then positivity and uniqueness follows by Theorem 2.2. $\square$

The following $\varepsilon-$compactness result is a basic tool in the proof of Theorem 1.3.

by Rellich Theorem. Applying (4.1), Hölder inequality and (4.2) we obtain

We estimate from below the left hand side of the above inequality by

Since $\varepsilon_{0}^{\frac{q-2}{q}}<S_{q}(\lambda)$ and $\phi\equiv 1$ on $B_{R^{\prime}}$, the conclusion follows by the arbitrariness of $R^{\prime}\in(0,R)$. $\square$

Using Ekeland’s variational principle we can fix a minimizing sequence for $S_{q}(\lambda)$ such that

which readily leads to a contradiction by Rellich theorem, as $q\in(2,2^{*\!*})$. Proposition $(i)$ in Theorem 1.3 is completely proved. $\square$

We start by pointing out a sufficient condition for existence.

If $S_{2^{*\!*}}(\lambda)<S^{*\!*}$ then $S_{2^{*\!*}}$ is achieved.

Proof. As in the proof of part (i) we select a minimizing sequence for $S_{2^{*\!*}}(\lambda)$ satisfying

Now assume $0<\lambda<\gamma_{N}^{2}$ and let $U$ be an extremal for $S^{*\!*}$. Then

and hence $S_{2^{*\!*}}(\lambda)$ is achieved by Lemma 4.2. $\square$

Positivity, symmetry and breaking symmetry

In this section we study the symmetry and breaking symmetry of $S_{q}(\lambda)$ depending on the parameter $\lambda$. We identify functions $u$ that coincide up to a multiplicative constant and up to a transform of the kind (1.7).

We start by recalling that in case $\lambda=0$ and $q=2^{*\!*}$, the Sobolev constant $S^{*\!*}=S_{2}^{*\!*}(0)$ is achieved by the radially symmetric function

Since truncations $u\mapsto u^{\pm}$ are not allowed in dealing with fourth order differential operators, the positivity of extremals for $S_{q}(\lambda)$ does not follow by standard arguments.

In the first result we use rearrangement techniques and the uniqueness result in Theorem 1.2 to investigate the case $\lambda\geq 0$.

In addition, since we are assuming that $u^{*}\neq u$, then

and that the strict inequality holds if $\lambda>0$. Similarly, we find

and the strict inequality holds if $\beta>0$, that is, if $q<2^{*\!*}$. In conclusion, since we are assuming that $\lambda$ and $\beta$ are not contemporarily zero, we have that

a contradiction. Therefore $u=u^{*}$, that is, $u$ is nonnegative and radially symmetric decreasing function.

Uniqueness follows immediately by using the Emden-Fowler transform as in Section 3 and Theorem 1.2. $\square$

As soon as $\lambda\to-\infty$, a braking symmetry phenomenon appears. The next theorem applies in case $q<2^{*\!*}$, due to the nonexistence result pointed out in the critical case $q=2^{*\!*}$, $\lambda<0$. We quote , and for remarkable breaking symmetry results for similar second-order equations.

If $\lambda<<0$ then $S_{q}(\lambda)<S_{q}^{\rm rad}(\lambda)$ and hence no extremal for $S_{q}(\lambda)$ is radially symmetric.

Proof. We already know that $S_{q}(\lambda)\leq S_{q}^{\rm rad}(\lambda)$. We will give an explicit condition on $\lambda$ to have $S_{q}(\lambda)<S_{q}^{\rm rad}(\lambda)$. Define

Define the test function $v$ as $v=u\varphi_{1}.$ Thus

by Hölder inequality. Therefore from (5.1) and from the definition of $Q(u)=n(u)$ we obtain

then no extremal for $S_{q}(\lambda)$ is radially symmetric and symmetry breaking occurs. $\square$

If $q$ is close enough to $2^{*\!*}$ then one can obtain a better estimate on the breaking symmetry parameter $\lambda$ by arguing as follows. Notice that $X>\gamma_{N}$ by the Rellich inequality (1.4). Thus, if $\gamma_{N}\geq\frac{N-1}{q-2}$ that is, if

then the radial solution $u$ does not achieve $S_{q}(\lambda)$ unless

Appendix A ​​​​​​ppendix

The following lemma has been used in the proof Theorem 5.1 (see also , Remark II.13).

Lemma A.1 can be proved by investigating the integrability properties of the Green’s function at infinity. However, we will use here an alternative approach that allows us to underline the unexpected role of the Rellich inequality.

for any $u\in C^{\infty}_{c}(\mathcal{C}_{\Sigma})$.

Proof. Fix any $u\in C^{\infty}_{c}(\mathcal{C}_{\Sigma})$. Use the divergence theorem and Proposition 1.1 in to estimate

If (A.2) holds, then in particular the Hardy constant in inequality (A.1) is positive, and we can define a Hilbert space $\mathcal{D}^{1,2}(\mathcal{C}_{\Sigma};|x|^{-2}~{}dx)$ of maps $u$ vanishing on $\partial\mathcal{C}_{\Sigma}$ (if not empty), and such that

Then $\mathcal{D}^{1,2}(\mathcal{C}_{\Sigma};|x|^{-2}~{}dx)$ is continuously embedded into $L^{2}(\mathcal{C}_{\Sigma};|x|^{-4}dx)$. Next we put

By the Rellich inequality on cones proved in and using (A.2) we get that

Thus $\mathcal{N}^{2}(\mathcal{C}_{\Sigma})$ is a Hilbert space with norm

and $\mathcal{N}^{2}(\mathcal{C}_{\Sigma})\hookrightarrow L^{2}(\mathcal{C}_{\Sigma};|x|^{-4}dx)$.

Moreover, $v$ satisfies the Navier boundary conditions

Proof. We only have to prove existence. Uniqueness easily follows. For any $R>1$ we put

Let $v_{R}\in H^{1}_{0}(A^{\Sigma}_{R})$ be the unique solution to

We denote by $v_{R}\in\mathcal{D}^{1,2}(\mathcal{C}_{\Sigma};|x|^{-4}dx)$ the null extension of $v_{R}$. Our aim is to use $|x|^{-2}v_{R}$ as test function in (A.3). Notice that

by the Cauchy-Schwarz inequality and by Lemma A.2. Since $\gamma_{N}+\lambda_{\Sigma}>0$ by assumption, we first infer that $v_{R}$ is bounded in $L^{2}(\mathcal{C}_{\Sigma};|x|^{-4}dx)$. Thus, using (A.4) again, we conclude that $v_{R}$ is uniformly bounded in $\mathcal{D}^{1,2}(\mathcal{C}_{\Sigma};|x|^{-2}~{}dx)$. Therefore, up to a sequence $R\to\infty$, we can assume that $v_{R}\rightharpoonup v$ weakly in $\mathcal{D}^{1,2}(\mathcal{C}_{\Sigma};|x|^{-2}~{}dx)$. It is easy to prove that $v\in\mathcal{D}^{1,2}(\mathcal{C}_{\Sigma};|x|^{-2}~{}dx)\hookrightarrow L^{2}(\mathcal{C}_{\Sigma};|x|^{-4}dx)$ solves $-\Delta v=g$ on $\mathcal{C}_{\Sigma}$. Thus in particular $\Delta v=-g\in L^{2}(\mathcal{C}_{\Sigma})$, that implies $v\in\mathcal{N}^{2}(\mathcal{C}_{\Sigma})$. In case $\partial\mathcal{C}_{\Sigma}$ is not empty, then $v$ satisfies Navier boundary conditions by standard arguments. $\square$

Acknowledgements. This research was done when the first author was visiting the ICTP Mathematics Section in Trieste, Italy, and travel grant was sponsored by National Board of Higher Mathematics (NBHM), India. Warm hospitality of ICTP is gratefully acknowledged.

The second Author wishes to thank Prof. Fabio Zanolin for many helpful discussions about equation (2.4) and for having suggested the references , .

References